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Math Analysis Homework - Week 2

Class 4

T1

x1>1,xn+1=3xn+1xn+3,n=1,2,

xn>11<xn+1=3xn+1xn+3<3Inductively,xn(1,3)xn+1xn=3xn+1xn23xnxn+3=1xn2xn+3<0{xn}is bounded and decreasinglimnxnexists.xn+1=3xn+1xn+3limnxn=3(limnxn)+1(limnxn)+3limnxn=1

T2

0<a1<b1, 令 an+1=anbn,bn+1=an+bn2,nN+. 证明: {an},{bn} 收敛于同一极限.

n,an<bn{bn>an+1=anbn>anan<bn+1=an+bn2<bn{an<bn<bn1<<b1,bn>an>an1>>a1{an},{bn} is bounded and monotonicA=limnan,B=limnbn exists{limnbn+1=limnan+bn2limnan+1=anbn{B=A+B2A=ABA=B

T3

设数列 {xn} 满足 0<xn<1(1xn)xn+1>14,n=1,2,3,, 求证: limnxn=12.

14(1xn)<xn+1<1xn<3414(1xn1)<xn<34xn1<23同理xn2<12n,xn<12xn+1=14(1xn)>xn(2xn1)2>0TrueX=limnxn existsxn+1>14(1xn)X14(1X)X=12

T4

x1(0,1),xn+1=xn(1xn),nN+, 证明: 数列 {nxn} 收敛, 并求其极限.

limn1xnn Stolz Theorem limn1xn1xn1=limn1xn1(1xn1)1xn1=limn11xn1=11limnxn1

对于x,

x111,x2<min(x1,1x1)12xn1n,n2xn+1<1n(11n)=n1n2<1n+1}xn1n0<xn1nSqueeze Theoremlimnxn=0limn1xnn=11limnxn=1limnnxn=1

T5

求极限 limn(n!e[n!e]).

an=n!e[n!e]=limn{n!e}=limn{i=0n!i!}=limn{i=n+1n!i!}=limn{1n+1+1(n+1)(n+2)+1(n+1)(n+2)(n+3)+}<limn{i=1(n+1)i}=limn{1n}=0

T6

求极限 limnan(1n+1+1n+2++1n+n).

x>ln(x)+11x(ln(xx1),ln(x+1x))an(i=n+12nln(ii1),i=n+12nln(i+1i))=(ln(2nn),ln(2n+1n+1))Squeeze Theoremlimnan=ln(2)

T7

设数列 xn=(1+12)(1+122)(1+12n), 证明 limnxn 存在.

xn is obviously incresingln(xn)=i=1nln(1+12i)<i=1n12i<1xn<elimnxnexists

Class 5

T1

用柯西收敛准则证明数列收敛.

an=i=2nsin(ix)i(i+sin(ix)),xR
|an+man|=|i=n+1n+msin(ix)i(i+sin(ix))|<i=n+1n+m|sin(ix)i(i+sin(ix))|<i=n+1n+m1i(i1)=i=n+1n+m1i11i=1n1n+m<1nϵ,N:=1ϵ+100i,j>N,|aiaj|<ϵCauchy Convergence TheoremQ.E.D

T2

bn=i=1n1|ai+1ai| is boundedan is convergent
bn is increasingbn is bounded}limnbn=Ban+man=i=n+1n+maiai1i=n+1n+m|aiai1|=bn+mbnϵ1,Ns.t.n>N|bnB|<ϵ1bn+mbn<|bnB|+|Bbn+m|=2ϵ1ϵ1:=ϵ2x,y>N,|axay|<ϵCauchy Convergence TheoremQ.E.D

T3

ϵ,N1=N(ϵ)s.t.i,j>N1|xixj|<ϵxn is convergent

逆向三角不等式显然.考虑正向

ϵ1:=1i>Nxi[xNϵ1,xN+ϵ1]a1:=xNϵ1,b1:=xN+ϵ1

[ai,bi],考虑取 ϵi=ϵi12,有Ni=max(Ni1,N(ϵi)) s.t. j>Ni,xj[xNiϵi,xNi+ϵi].

于是令ai=max(ai1,xNiϵ),bi=min(bi1,xNi+ϵi). 显然|biai|<12i1,于是

aiai+1bibi+1limnbnan=0}!ξ[an,bn]s.t.ϵ,N=min{i|biai<ϵ}n>N|ξxn|<ϵQ.E.D

T4

a0=3,an=an122{limnan=+An:=ani=0n1ailimnAn=5

(1)

a0=3,a1=7,a2=47n>2,an1>2nan=an122>22n2>2n+1}inductionan>2n+1limnanlimn2n+1=

(2)

an=an122(an2)=(an12)(an1+2)i=1n(ai+2)=an+12a12=an+125An2=an2i=0n1ai2=an+1+2i=1n(ai+2)=5an+1+2an+12limnAn=5an+1+2an+12=5

Class 6

T1

A,Bis upper bounder,S{x+y|xA,yB}supSsupA+supB

显然S有界,则supS存在.

supS>supA+supB,取M=supA+supB,

xA,yB,xsupA,ysupBx+yM

于是M是比supS小的上界,矛盾.

supSsupA+supB

eg. 当A,B,S均为有限集恰好为 ${ x+y\vert x\in A,y\in B } $ 时显然严格不等号不成立.

T2

A,B aren’t empty,α0,C=A+αB={x|x=a+αb,aA,bB}}sup(A+αB)=M=supA+αsupB
cC=a+αb,asupA,bsupBa+αbsupA+αsupB=MM<M,ϵ=MMlet X=supAϵ3,Y=supB+ϵ3αAccording to the definition of supremum, a>XA,y>YBc=a+bC,c>X+Y>MsupC=supA+αsupB

T3

ABsupAsupB,infAinfB
if supA>supBlet M=supBDef of supremumaA,a>M=supBaAABaBa>supB}FalsesupAsupB

C=A,D=B,CD,supCsupDinfAinfB,

T4

xA,yB,xysupAinfB

假设supA>infB,M(supA,infB),由supA定义aA,a>M,由infB定义bB,b<M,故a>M>b,与xA,yB,xy矛盾.

supAinfB

Linear Algebra Done Right

记录大致讲了什么

向量空间

我们定义向量空间

向量空间

向量空间定义在域F上,要求支持:

  • 加法
  • F中数的数乘
  • 加法单位元(0元)
  • 加法逆元
  • 加法交换结合
  • 数乘结合
  • 加法对数乘分配
  • 对加法和数乘封闭

加粗部分用于判断子空间,根据F区分是否是实向量空间/复向量空间

子空间

就是子集且是向量空间的对吧

判定可以看上面

维数

向量空间还应该有维数.于是定义

线性无关组

Groupv1vnVis linear independet{cn},ciF,i=1ncivi=0i,ci=0

张成,张成组

span(v1vn)={i=1ncivi}

张成V的组简称张成组

基=张成组+线性无关组

则按照直觉的,维数应该是张成组长度的最小值,线性无关组长度的最大值,基的长度等等,下面讨论的是有限维线性空间.

  • dimV是张成组最小长度
  • dimV是线性无关组组最大长度
  • dimV是任意一组基的长度

首先说明,对一个线性相关的组,我们一定可以去掉一个线性相关项保持张成空间不变(显然).

首先说明

任意一组线性无关组长度小于等于任意一组张成组长度:

考虑一个线性无关组和一个张成组,将一个线性无关组的元素加入张成组,则形成的一定是线性相关组,删去一个张成组中的元素则保持张成,不断重复这个操作,注意因为被加入的线性无关,所以你想相关一定得带张成组中的,于是可以一直操作.

直到线性无关组全部被加入,则因为每次删掉一个张成组元素,你一定有线性无关组长度不大于它.

线性无关组可通过加元素扩展到基,张成组可通过删除元素到基

对线性无关组,每次加入一个不属于张成空间的,由于组的长度大小不断增加,而存在一个长度的基,所以你的过程会停止.

同理对张成组每次删掉一个线性相关项不影响张成空间.

每组基的长度都相当,具有恰当长度的线性无关组/张成组是基.

第一句用小于等于关系,后面两个通过基的长度相等+上一条可以变成基说明.

维数就是这个长度为 dimV

空间的运算

空间的和

U+V={u+v|uU,vV}

加法实际上是并(包含子空间所有向量的最小空间).

并定义直和.

空间直和

W=UV=U+Vs.t.wW,\exist!uU,vV,u+v=w

直和相当于不交并,是对空间进行一种分解.

然后直和的判定容易证明只要0的表示满足唯一性,并推出当且仅当它们的交只有0

空间的积

就是笛卡尔积.

U×V={(u,v)|uU,vV}

其实也可以先扩展一下直和 是等价的.

仿射空间

u+V={u+v|u+V}(u+V)+(w+V)=(u+w)+Vλ(u+V)=λu+V

注意如果uV则不变,uV则是一个没有0的空间,在此空间的线性组合变成了i=0ncivi s.t. i=0nci=1.

然后仿射空间关于以上运算是线性空间,其中0=V

空间的商

U/V={u+V|uU}

可以理解成等价类分类.如果两个向量的差在V中则认为他们等价.新的空间中每个元素都是一个等价类.而你仿射空间的变换也可以看成是对代表元做变换.

最后考虑它们的维数,有:

dimU×V=dimUV=dimU+dimVdimU+V=dimU+dimVdimUVdimU/V=dimUdimV

第一行是显然的.

第二行考虑取UV的一组基,再其中添加上UV的基和VU的基.

第三行需要线性映射.

线性映射

基本性质

线性映射

线性映射是映射满足:

  • 齐性: Tλv=λTv
  • 加性: T(u+v)=Tu+Tv

线性变换的运算

(S+T)u=Su+Tu(λT)u=λ(Tu)(ST)u=S(T(u))

符合直觉的.于是你可以说明L(U,V)(由T:UV组成的集合)是 dimU×dimV维的线性空间,他的标准基可以是所有把U的一个基映到V的一个基的映射.

值域,零空间

for linear map T:UVrange T={v|v=Tu,uU}null T={u|Tu=0,uU}
dimU=dimnull T+dimrange T

考虑 null T 的基u1un,并添加v1vm扩充到U的基.

对任意x=iciui+icivi,我们有Tx=iTcivi=iciTvi,故任意 wrange T可以表示成Tvi的线性组合,且显然Tvi之间线性无关,于是就得证.

单射,满射,双射,可逆

  • 单射:TxTyxy
  • 满射: vV,uU,Tu=v
  • 双射就是同时有两条
  • 对于双射T的定义T1满足TT1=T1T=I

基本性质

  1. 单射等价于只把0映射到0.
  2. 存在UV单射说明 dimUdimV.
  3. UV有单射说明VU有满射.
  4. UV存在双射称为U,V同构,可以证明任意向量空间同构于某个Fn.

都挺显然的.

dimU/V=dimUdimV

定义 TL(U,U/V) 为商变换,把u映射到u+V.

null T=Vrange T=U/V

套用上面值域和零空间的维数公式即可.

矩阵

矩阵

选取U,V分别一组基u1um,v1vn,可以把线性变换T写成n×m的矩阵M(T,u1um,v1vn)=An×m满足

Tui=j=1nAj,ivj

即每一列是一个基向量在像空间中的基的表示.然后有的时候也直接简写M(T).

矩阵可以这么定义因为线性变换的线性保证了你可以只用基的变换去描述它,同时基的这种变换也能唯一确定线性变换.

于是可以定义矩阵运算:

M(S)M(T)=M(ST)M(S)+M(T)=M(ST)λM(T)=M(λT)

其中第一行矩阵乘法用坐标写一下可以推出经典的矩阵乘法方式.

对偶

线性泛函,对偶空间

线性泛函就是fL(V,F),所有这样的f组成线线性空间VV的对偶空间.

线性泛函可以看成是向量/点的对偶.线性泛函 {φ|φiej=[i=j]}构成对偶空间的基.

对偶映射

TL(U,V) ,定义 TL(V,U) 满足

fV,Tf=fT

T是反的可以理解因为V的泛函的输入才是T的输出.导致没法根据这几个东西定义一个正的出来.

然后对偶主要解释了:M(T)=M(T)T(右上角的T是转置的意思).

对偶映射的运算

(ST)=TS(S+T)=S+T(λS)=λS
(ST)f=fST=T(fS)=T(Sf)(S+T)f=f(S+T)=fS+fT=Sf+Tf(λS)f=fλS=λfS=Sf

零化子

对线性空间V来说,子空间U的零化子 U0={f|fV,uU,fu=0}.

注意U0同时依赖UV.

dimU+dimU0=dimV

U的基u1un扩充到V的基u1un,un+1un+m.并取V的标准基φiuj=[i=j],则显然fU0要求f不能有φi,i<n的分量,而任意i>n的分量都可以有.于是得证.

null T=(range T)0range T=(null T)0

考虑Tfu=fTu=0关于所有u成立,则f的范围是什么.看右侧显然是 (range T)0 看左侧则是 (null T)0.于是得证.

对第二行,左边是任意Tg=gT,右边说你这个线性泛函把所有Tu=0的映到0,恰好是左边的gT满足条件.于是得证.

然后还有一个问题是我们以为T=T,但实际上你甚至不能保证VV是相同的.然后有个典范同构的概念形容他俩的关系就是存在一种不依赖于基的选取的同构(只要定义T(u)f=fu,则uT(u)是双射.)

本征值

本征值,本征向量

若对算子T,v0V,λF s.t. Tv=λv,则λ,v分别为本征值,本征向量.

就是说算子在这个方向上对变换只有伸缩.

一个本征值可能对应多个线性不相关的本征向量,它们构成本征空间E(T,λ)

λT的本征值等价于 TλI不是双射,或不是单射/满射

首先注意到对算子来说 单射,满射双射等价

又因为单射等价于 null T=0 所以 (TλI)v=0 和它不是单射等价.

不同本征值对应对本征向量线性不相关.

反证,你要利用不同本征值这个性质,于是你设 vnspan(v1vn1)n为满足条件对最小的.

vn=i=1n1civiTvn=i=1n1Tciviλn(i=1n1civi)=i=1n1λicivi0=i=1n1(λiλn)civi

因为n是最小的,所以v1vn1线性无关,然后你就推出矛盾.

有此容易说明本征值个数不大于线性空间维数.

复向量空间中的线性映射一定有本征值

考虑

vV,dimV=nv,Tv,T2vTnV is dependenti=0nciTiv=0代数基本定理(i=1n(TλiI))v=0i,TλiI=0λiis a eigenvalue of T

不变子空间,商算子和限制算子

不变子空间

即对算子T,有uU,TuU,则U为不变子空间.

商算子,限制算子

T/UL(V/U,V/U),(T/U)(v+U)=(Tv+U)T|UL(U,U),T|Uv=Tv

显然T|U要求了U是不变子空间.

把算子放到更小的空间去研究的方式.

上三角矩阵

按照上面基的理解,有

M(T,u1un)is upper triangular matrixi,Tuispan(u1ui)i,span(u1ui)is invariant space

感觉是显然的.

那么考虑什么样的线性映射T有一组基u1un有上三角矩阵An×n.

TL(V,V),V is complex vector spaceu1un,M(T,u1un) is upper triangular matrix

Proof 1

归纳,假设对任意维数小于dimV的空间成立,考虑取T的任意本征值λ,则 U:=range TλI,则因为T不是单的所以 dimU<dimV.且 uU,Tu=(TλI)u+λuU,所以TU不变.

于是可以应用归纳结假设,T|UU上有一组基u1un使得 M(T|U,u1un) 是上三角矩阵.

将这组基扩展到V上成为u1un,v1vm,则对任意 w=i=1nui+i=1mvi,Tv=

Proof 2

对角矩阵

English Speech

Class 1 essay

Generative AI Enhances Human Creativity

To begin with my argument, I’ll tell a story of my own. In the summer vacation, with generative AI I developed a online notebook application whose design may be innovative and unique in my view. To complete this work I, a high school student, used AI to generative tons of code. People’s creativity comes from their whims, and most of them just evaporate for their limited capbility. However, when AI’s knowledge and skills shelter the whims from the complexity of the reality, our whims could transform into a true, creativity, rather than fade out in our mind. This is the first point.

Secondly, AI proivdes a more accessible platform for every one to gain knowledge. We can’t deny that creativity is based on knowledge. Primitive human can’t imagine the Internet while the under-educated people can’t understand generative AI’s structure as well. Today ,those difficult conception could be explained by AI, those hidden tips could be unearthed from the corner of the Internet by AI, and those people, who consist the most part of the world and can’t accept the best education, could have the most patient teacher. So AI will definitely enhance creativity by expanding the soil it comes from.

Lastly, if we omit the content I had explained and stubbornly insist AI hinder the creativity, saying like “AI’s convenient answer prevent people from thinking” or some prediction about a dark future with AI, I will say that it’s not AI hinder our creativity. When employers replace workers with ai without considering their lives, when some people depending on ai give up thinking deeply, the faughts were always on people ourselves. It may be inappropriate structure of society or the laziness of human nature. AI is just a knife and people decide who it will attack.

As a summry, AI has enhanced and will constantly enhance human’s creativity by realizing the whims that we could already quit and popularizing the knowledge be the basical root of creativity.

Linear Algebra Class1 Homework

T2

Going around a triangle from (0,0) to (5,0) to (0,12) to (0,0), what are those three vectors u,v,w? What is u+v+w? What are their lengths u, v, and w? The length squared of a vector u=(u1,u2) is u2=u12+u22.

{u=[50]v=[512]w=[012]u+v+w=0||u||=5||v||=13||w||=12

T3

Describe geometrically (line, plane, or all of R3) all linear combinations of

  • (a) [123] and [369]
  • (b) [100] and [023]
  • © [200] and [022] and [223]
(a):

a line with direction vector [123]

(b):

a plane through the origin

©:

all the 3d space

T5

If v+w=[51] and vw=[15], compute and draw the vectors v and w.

v=[33]w=[22]

T7

Compute u+v+w and 2u+2v+w. How do you know u,v,w lie in a plane?

u=[123]v=[312]w=[231]

These lie in a plane because w=cu+dv. Find c and d.

u+v+w=02u+2v+w=[231]{c=1d=1

T11

If three corners of a parallelogram are (1,1), (4,2), and (1,3), what are all three of the possible fourth corners? Draw those three parallelograms.

(4,0),(4,4),(2,2)

a triangle with another three outer vertices

T13

Review Question. In xyz space, where is the plane of all linear combinations of i=(1,0,0) and i+j=(1,1,0)?

equal to the linear combination of i and j(j=(0,1,0)).

so it’s the plane xOy

T17

What combination c[12]+d[31] produces [148]? Express this question as two equations for the coefficients c and d in the linear combination.

{14=1c+3d8=2c+1d{c=2d=4

T19

Restricted only by c0 and d0 draw the “cone” of all combinations cu+dv.

a angle that has the origin as the vertex(we should paint all the points between u and v black)

T23

If you look at all combinations of those u,v,w, is there any vector that can’t be produced from cu+dv+ew? Different answer if u,v,w are all in ______.

(1):no.

(2):a plane

Challange Problems

T24

How many corners (±1,±1,±1,±1) does a cube of side 2 have in 4 dimensions? What is its volume? How many 3D faces? How many edges? Find one edge.

(1):every component of every vertex’s coordinate has two value, so there are 24=16 vertices.

(2):volume: 24=16

(3):every 3d faces means a restriction of one component, so 42=8

(4):every edge means a restriction of three components, and the last component’s two value represents two vertex it connects. so (43)23=32

T25

Find two different combinations of the three vectors u=(1,3) and v=(2,7) and w=(1,5) that produce b=(0,1). Slightly delicate question: If I take any three vectors u,v,w in the plane, will there always be two different combinations that produce b=(0,1)?

(1)

w=3u+2v,b=2u+2v.

so for each t

{u=23tv=2+2tw=t

is a solution

(2)

No. eg: u=v=w=(1,0)

Yes when there are two vectors in {u,v,w} that are linearly dependent.

T26

The linear combinations of v=(a,b) and w=(c,d) fill the plane unless ________. Find four vectors u,v,w,z with four nonzero components each so that their combinations cu+dv+ew+fz produce all vectors in four-dimensional space.

(1)

k s.t. w=kv or u=0 or v=0

(2)
{u=[1,1,1,2]v=[1,1,2,1]w=[1,2,1,1]z=[1,1,1,1]

(do some elementary row operation on Identity Matrix)

T27

Write down three equations for c,d,e so that cu+dv+ew=b. Write this also as a matrix equation Ax=b. Can you somehow find c,d,e for this b?
u=[210]v=[121]w=[012]b=[100]

{c=34d=12e=14

Math Analysis Homework - Week 1

Class 1 Homework

T1

anbncn,limn(cnan)=0an收敛

Obviously wrong.

an=bn=cn=n

T2

anbncn,bn收敛,limn(cnan)=0an收敛
ϵ1>0,N1 s.t. n>N1cnan<ϵ1bnancnan<ϵ1ϵ2>0N2 s.t. n>N2bnB<ϵ2|anB||bnan|+|bnB|ϵ+ϵ2ϵ1,ϵ2:=ϵ2ϵ,N:=max(N1,N2),n>N|anB|<ϵQ.E.D

T3

liman=A,an0liman+1an=1

Wrong

an=2n

T4

limnanbn=0(limnan)(limnbn)=0

Wrong

an=(nmod2)bn=((n+1)mod2)

T5

limnbnan=1,limnan=Alimnbn=A

不妨设A>0,又因为取ϵ<A可以让n>Nan>0,故不妨设an>0

,ϵ1<1,ϵ2<A

ϵ1N1 s.t. |anbn1|<ϵ1ϵ2N2 s.t. |anA|<ϵ2anbn(1ϵ1,1+ϵ1)an(Aϵ2,A+ϵ2)bn=ananbn(Aϵ21+ϵ1,A+ϵ21ϵ1)bnA(Aϵ1ϵ21+ϵ1,Aϵ1+ϵ21ϵ1),|bnA|Aϵ1+ϵ21ϵ1<ϵϵ1:=ϵ100A,ϵ2:=ϵ100|bnA|Aϵ1+ϵ21ϵ1=ϵ501ϵ100Alet ϵ<A1ϵ100A>150N=max(N1,N2)n>A|bnA|<ϵQ.E.D

T6

limn3n2+n2n21=32
3n2+n2n2132>3n22n232=03n2+n2n2132=3n2+n32(2n21)2n21=n+322n21<n>12nn2=2n<ϵN:=2ϵ+114514Q.E.D

T7

limnn2+nn=12
n2+nn=(n2+nn)(n2+n+n)n2+n+n=nn2+n+n=11+1+1n|11+1+1n12|=1211+1+1n=1+1n12(1+1+1n)<1+1n1<Bernoulli Inequality1+12n1=12nN:=12ϵ+100Q.E.D

T8

n2arctan(n)1+n2=π2
n2arctan(n)1+n2=n21+n2arctan(n)
limnn21+n2=1
lim|n21+n21|=11+n2<1nN:=1ϵ+100Q.E.D
limnarctan(n)=π2
N:=tan(π2ϵ2)ϵ,|arctan(n)π2|=π2arctan(n)=ϵ2<ϵQ.E.D
(liman)(limnbn)=X,an>0,bn>0limnanbn=X
A:=liman,B:=limbnϵ1,N1 s.t. n>N1|anA|<ϵϵ2,N2 s.t. n>N2|bnB|<ϵn>max(N1,N2)an(Aϵ1,A+ϵ1),bn(Bϵ2,B+ϵ2)anbn((Aϵ1)(Bϵ2),(A+ϵ1)(B+ϵ2))|anbnAB|<Aϵ2+Bϵ1+ϵ1ϵ2ϵ,ϵ2:=ϵ4A,ϵ1:=ϵ4B|anbnAB|=ϵ2+ϵ216AB<ϵ<ABϵQ.E.D
limnn2arctan(n)1+n2=limnn21+n2limnarctan(n)=1×π2=π2Q.E.D

Class 2

T1

limn(2)n+3n(2)n+1+3n+1

limn(2)n+3n(2)n+1+3n+1limn=13(23)n+1(23)n+1+1limn=13

T2

limn[112+123++1n(n+1)]

=limni=1n1i(i+1)=limni=1n1i1i+1=limn11n+1=1

T3

limn[1n2+1+1n2+2++1n2+n]

i=1n1n2+1i=1n1n2+ii=1n1n2+nSqueeze Theoremlimni=1n1n2+1L=limni=1n1n2+ilimni=1n1n2+nlimnnn2+1Llimnnn2+nlimn11+1n2Llimn11+1nL=1

T4

limnn2n+2n

Obviously: n>2n2n+2>8>1

|(n2n+2)1n1|=(n2n+2)1n1<ϵn2n+2<(1+ϵ)nF(x)=n2n+2<1+nϵ+(n2n)ϵ22+n(n1)(n2)6ϵ3=G(x)C1(ϵ),C2(ϵ)s.t.F(x)<C1(ϵ)n2,G(x)>C2(ϵ)n3N:=C1(ϵ)C2(ϵ)+1(n>N(n2n+2)1n1<ϵ)limnn2n+2n=1

T5

limnarctan(n)n
1<arctan(n)<π21<arctan(n)n<π2n1<arctan(n)n<π2nlimn1=limnπ2n}Squeeze Theoremlimnarctan(n)n=1

T6

limn2sin2(n)+cos2(n)n

同上一题,里面有界,是1

T7

limn[nan]n, 这里 limnan=a.

nan1n<[nan]n<nan+1nlimnnan+1n=limnan+limn1n=alimnnan1n=limnanlimn1n=a}Squeeze Theoremlimn[nan]n=an

T8

证明 an=2n+(1)nn3n+1 发散

limna2n=limn6n6n+1=1limna2n+1=limn2n+16n+4=131}{an}发散

T9

an0,an+1an>0,limnan+1an=0N s.t. n>N{an}单调
use ϵ1=1,N1s.t.limnan+1an<1exact an+1<an

T10

limn(xnxn1)=d, 证明: limnxnn=d.

ϵ1,N1 s.t. n>N1|xnxn1d|<ϵ1xnxn1[dϵ1,d+ϵ1]xn=xN1+i=N1+1n(xixi1)[xN1+(nN1)(dϵ1),xN1+(nN1)(d+ϵ1)]xnn[xN1N1(dϵ1)n+dϵ1,xN1N1(d+ϵ1)n+d+ϵ1]ϵ,xnnd[xN1N1(dϵ1)nϵ1,xN1N1(d+ϵ1)n+ϵ1]ϵ1:=ϵ2,n:=2xN1N1(dϵ1)ϵ|xnnd|<ϵ

T11

limnan=a(a>0,nN), 证明: limna1a2ann=a, 并由此证明:

  • limnan+1an=a(a>0,nN),则 limnann=a;
  • limnn!nn=1e

(1)

Solution 1
|iaina|<ϵaϵ<iain<a+ϵ(aϵ)n<iai<(a+ϵ)nϵ1,N1 s.t. n>N1an[aϵ1,a+ϵ1]iai=(i=1N1ai)(i=N1+1nai)[A(aϵ1)nN1,A(a+ϵ1)nN1]letϵ1:=ϵ2,Consider A(aϵ1)nN1>(aϵ)n:A(aϵ1)N1>(aϵaϵ1)nSince aϵaϵ1<1,N2 s.t. n>N2不等式成立右侧同理有N3N:=N1+N2+N3s.t.n>N(aϵ)n<A(aϵ1)nN1<iai<A(a+ϵ1)nN1<(a+ϵ)n|iaina|<ϵQ.E.D
Solution 2

唐. 可以用调和均值/算数均值夹两边.

(2)

bn=an+1an,则问题转化为(1).

(3)

e=limn(1+1n)n1e=limn(nn+1)nan:=(nn+1)n检验符合引理Q.E.D

T12

limnxn=+, 证明: limnx1+x2++xnn=+.

X,X1:=X+1limnxn=N1 s.t. n>N1xn>X1=X+1for n>N1,i=1nxin=i=1N1xi+i=N1+1nX+1n>(1N1n)(X+1)n:=N1(X+1)+100i=1nxin>X

T13

an>0,limnanan+1+an+2=0anis unbounded

反证,设M0<an<M.

ϵ=15,anan+1+an+2<15an+1+an+2>5anmax({an+1,an+2})>2anb1=N+1,bi=abi1+1,abi1+2中较大的一个的下标abi>2iab1

a有发散子列,a发散.

T14

设数列{xn}单调增加, limnx1+x2++xnn=a, 证明: limnxn=a.

若存在xN>a,则n>Nxn>a,limnxN>a,则A(xN,A). 并有iN,xi>A>a,则

limni=1nxin>limni=1Nain+nNn(a+(Aa))=A>a

矛盾,故i,xi<a

于是

i=1nxin<xn<alimna=alimni=1nxin=a}limnxn=a

Class 3

Math Analysis Homework - Class 3

T1

limn1+12++1nlnn

limn1+12++1nlnnStolz Theoremlimn2nlnnn1>x>1ln(x)>2x1x+1limn2n22n1=+

T2

limn1+2+33++nnn

limn1+2+33++nnnStolz Theoremlimnnn1=1

T3

limna1+2a2++nani=1ni (已知 limnan=a)

limna1+2a2++nani=1niStolz Theoremlimnnann=a

T4

计算极限 limn(n!)1n2.

1<(n!)1n2<(nn)1n2=n1nlimn1=limnn1n=1}Squeeze Theoremlimn(n!)1n2=1

T5

xn=1n2k=0nln(nk), n=1,2,, 求极限 limnxn.

1n2k=0nln(nk)Stolz Theoremi=0nln(ni)i=0n1ln(n1i)2n1=i=0n1ln(nni)2n1=ln(nnn!)2n1=nln(nn!n)2n1According to homework class-2:limnn!nn=1elimnnln(nn!n)2n1=limnn2n1limnln(nn!n)=12

还是用了连续性/kk

T6

limnn(AnAn1)=0, 试证: 当极限 limnA1+A2++Ann 存在时, limnAn=limnA1+A2++Ann.

xn:=i=nnΔAii=i=1ni(AiAi1)=nAni=1n1AilimnxnnStolz Theoremxnxn1n(n1)=nΔAn=0limnAnlimni=1n1Ain1=limnxnn=0limnAn=a

Math analysis

Class 1 Some Inequality and Def of limit

Triangle Inequality

a,bR||a||b|||a+b||a|+|b|
|a||b|ab|a||b||a|22|a||b|+|a+b|2a2+2ab+b2|a|2+2|a||b|+|a+b|2(|a||b|)2(a+b)2(|a|+|b|)2

向量的模长也满足 也可以用这个做法.

Cauchy-Schwarz Inequality

0<xiRixinixinni1xi

显然 取xi=1xi能用左边不等式推右边 只证左边

归纳,2的幂次显然吧

对非2的幂次n,用X=ixin补完,直接套

Array’s Limit

limnan=Aϵ>0N s.t. nN|anA|<ϵ

注意

  • N=N(ϵ)
  • N 不唯一
  • ϵ 可限制在任意(0,a),a>0

Eg 1

limnn1n=1

|n1n1|=|(nn)1n1|=|(i1nn)1n1||n2+2nn1|2n

Or try this:

|n1n1|=n1n1<ϵn1n<(ϵ+1)n<(1+ϵ)nn<1+ϵn+ϵ2n(n1)2

Solve the equation, it’s a parabola with upward opening so the solution exists.

[think] At this stage, we cannot say limg(f(n))=g(limf(n)) which depends on continuity of function. But you sometimes can rewrite the proof with inequality.

Class 2

About Q

nZQC

p2=q2n Integer factorization

More natural than textbook but it doesn’t depend on Integer factorization.

pq is finite decimal or repeating decimal

Simulate how the division is done and consider the remainder will be repeated.

About Cardinality

A Example

Find a bijection of [0,1] and (0,1)

f(x):(0,1)[0,1]={0,x=121,x=131n2,x=1nx,otherwise

Wow!

Cardinality is comparable

A,B, f(x):ABs.t.f is injective or surjective

(承认选择公理)

首先选择公理出良序定理,找到A,B的良序(A,<),(B,<).

然后进行超限归纳,把A,B按照良序(保证了每个元素存在序里的后继),最小元素匹配,然后剩下的次小元素匹配,这么做下去.

然后这个看起来进行的是普通的自然数归纳显然是错的. 你需要用超限归纳也就是在序上做归纳.

然后问题来到序是什么.

序的结构是这样的 首先是自然数 自然数定义是0定义为空集开始 然后定义succ(S)=S{S}

然后定义完所有自然数后 定义ω=ii,就是把所有的自然数的集合并起来(显然它包含所有的自然数). 然后我们可以接下来用后继的定于去定义ω+1,ω+2,并且你又可以把它们并起来得到ω×2,不断走后继,ω,2ω,3ω可以变成ω2,又有ω3ωω等等

基本上是每个层次的运算完了之后进下一个层次的序数构造 总之它看起来包含了各种各样的无穷,可以应付所有大小的集合.

然后我们要在序数的结构上做归纳法,就要证明:xsucc(x),还要证明极限这一把也对,就是a1aniai这个操作(也就是对某个极限序数,如果所有它以前的序数推到它自己)合法,就满足了你可以不断到下一极限.这样推出对全体元素合法. 这就是超限归纳法.

那你对照一下我们的归纳就是一一对应啊,所以是合法的. 就结束了.

然后我们刚才是说明了可以借良序去给两个集合配对,那么你就一定能找到一个对另一个的单射,所以一定可比.

Cantor-Bernstein-Schröder Theorem

Card(A)Card(B),Card(B)Card(A)Card(A)=Card(B)

或者表达为

A,B,g:AB,f:BAf,g is injective}h:AB is bijective

考虑从任意元素uA可以引出一条链:uf(u)g(f(u)f(g(f(u))).
同理vB开始的链vg(v)f(g(v))g(f(g(v))).

注意到所有元素一定都在某条链上(映射).每个点度数一定至多一进一出(单射).

于是每条链上构造一个双射(显然的)拼起来即可.

Q.E.D

其实这些应该算集合论还是什么?

Class 3 Properties of Limit

  • Uniqueness:Obviously
  • Local boundedness: 取ϵ=1N,N后面显然,前面有限项也显然.
  • About subsequence:
    • about any subsequence
    • about some subsequence whose union is the sequence

Limitaion’s Calculation

limnan=A,limnbn=B
limnanbn=AB
|anbnAB|=|anbnanB+anBAB||aN||bnB|+|B||anA|

显然an,bn有界,令M是他俩共同的界,取ϵ=1114514M+114514用到an,bn上即证.

limnanbn=AB

先上一个保号性干掉分母上出现0的事.

证取倒数:

|1bn1B|<ϵ|Bbn|<ϵbnBϵ=ϵBM+114514(|M|>bn)Q.E.D

那转化到乘法是显然的.

无穷小

基本就是收敛到0的数列啊.

Eg

{limnan=alimnbn=blimniaibnin=ab

Solution 1

b=0时,考虑aM.

limniaibninlimnMbnin=0

b0,bi=bib,有

limniaibnin=limniaibnin+iainb=0+ab

Solution 2

不妨设a,b>0,则有界性可以得到后面是an,bn>0

考虑影响值的肯定是中间的项,所以直接拆,取ϵ,对a,b可以得到N1

limniaibnin==limni=1N1aibni+i=1N1biani+i=N1+1nN11aibnin=limni=1N1aibnin+limni=1N1bianin+limni=N1+1nN11aibnin(limnn2N11n(aϵ)(bϵ),limnn2N11n(a+ϵ)(b+ϵ))=((aϵ)(bϵ),(a+ϵ)(b+ϵ))

后面显然.`

Class 4

eg1

a>1,kN,limnnkan=0

an=(1+b)n展开会出现n的任意次方,然后显然.

Stolz

yn,limnyn=limnxnxn1ynyn1=a[,+]}limnxnyn=a

先简化,不妨设a0,ΔxnΔyn>0

然后可以证a=0时,你取ϵ1N1,变成 ΔxnΔyn<ϵ.于是

xnyn=xN1+i=N1+1nΔxiyN1+i=N1+1nΔyixN1+ϵi=N1+1nΔyiyN1+i=N1+1nΔyixN1+ϵi=N1+1nΔyii=N1+1nΔyi

因为你 Y=i=N1+1nΔyi+,所以一定可以有ϵY>xN1.就能证<2ϵ,你再取一下ϵ就得证了.

然后我说0<aR时你直接x=xay就用结论,a=+的时候你取倒数用结论,就做完了.

Class 5

单调有界数列有极限

ai>ai1ai<M}limnai exists

{ai}上确界M0,则任意ϵ,取M=M0ϵ,由确界知i s.t. ai(M,M0],则n>i,|anM|<ϵ.

然后书上因为没教你上确界,试图用无限小数说明.那其实就是你一位一位考虑,就先这一位增加到最大,然后进入下一位,容易发现最后区间长度趋近于0

其实无限小数就是区间套啊.

e

an:=limn(1+1n)n=bn:=limni=0n1i!

右边那个单调有界都是好证的.证相等:

an=(1+1n)n=i=0n1ni(ni)=i=0n1i!j=0i1(1jn)<bn

又有

k<nan=i=0n1i!j=0i1(1jn)>i=0k1i!j=0i1(1jn)

于是先令k得到an>bk,则a逐项大于b的一个子列,最后得证.

重点是把一次到极限拆成两个取极限.但这个为什么对呢.

是好证的,其实就是

[think]

limalimbf(a,b)=Xlimnf(n,n)=X

用定义是显然的.且如果极限都存在(要f(a,n)f(n,b)极限存在)的话你还可以看出左式两取极限交换(都是右式)

0<ei=0n1i!<1nn!
bn:=i=0n1i!X=bn+mbn=i=n+1n+m1i!<1(n+1)!i=0m11(n+1)i<1n!nebn=limmX<1n!n

然后说为什么取极限小于号还是小于号呢?

X=bn+mbn<bn+m+kbn<1n!nlimkebn<1n!n

[think] 对单调数列让一个独立变量趋近无穷说明严格不等号.

有个魔怔法,你先去下一条证明e是无理数再回来说取不了等.

eQ

假设e=pq

pqbn<1nn!n=qp(q1)!bqq!<1q

左边是不为0的整数(为0b增矛盾),右边是分数

[think] 一个无理数等价于存在一列分母到无穷的有理数始终有更高阶(相比序列中分母)的余项.

(1+1n)n(1+1n)n+1
(1+1n)n1<(n(1+1n+1)n+1)n+1=(1+1n+1)n+1

第二个取倒数同理.

从这个出发可以说明ln的切线放缩系列不等式.

有界数列必有单调子列

有界数列有收敛子列

可以考虑后缀max,取出一个单调子列.

也可以区间套,每次进有无穷多项的区间.

Class 6

Cauchy Convergance Theorem

Cauchy Convergance Theorem

ϵ,Ns.t.n,m>N,|anam|<ϵlimnan exists

右推左是显然的.

首先容易得到有界.于是它有收敛子列.

然后你其他的项到你的收敛子列的距离拿柯西的条件放缩一下就证完了.

或者也可以闭区间套.

确界原理

有上界的数集一定有上确界.

来闭区间套,二等分,如果上面的(包含边界)有就取上面,否则取下面,框出一个数.

然后来看,比他小的不是上界是好说的(取个区间即可).怎么说明它是上界呢?

考虑任何一个数,递归后一定有某一次它在下半区间(包含中点),那就证完了.

可能甚至不如无穷小数简洁 反正本质相同.

有限覆盖定理

S={(x,y)|x<y},TS,ITI[a,b]AT,IAI[a,b],|A|N(not infinity)

反证,设[A,B]不能被有限覆盖.

闭区间套 二等分 则一定有一半区间也是不能被有限覆盖的.递归到不能被有限覆盖的区间,最后弄出一个数.

但包含这个数的极小区间显然可以被有限覆盖.矛盾,得证.

很棒的啊,它完全不关心你无穷覆盖的结构而是到数的结构去了.

Class Unknown

Continuous function’s maximum/minimum

{x[a,b],f(x)<MM<M,x s.t. f(x)>Mf is continuousx,f(x)=M

考虑取任意M1<M,可以得到一个f(x1)(M1,M),取Mn>f(xn1)可得xn(Mn,M),显然xn有界可以取收敛子列yn,则limnf(yn)=f(limnyn),则于是得证

Intermediate value theorem

{x[a,b],f is continuousy[f(a),f(b)]x0,f(x0)=y

不妨设f(a)yf(b),=情况显然,只考虑f(a)<y<f(b).

A={x|f(x)<y} 则它有上确界x1.

那么一定存在一个收敛到x1的数列 {zn} 你就直接发现 limnf(zn)=f(limnzn)

Q.E.D

About Periodicty

{f(x) is a periodic function that isn’t constantf(x) is continuousf(x) has min positive period
T:={t|t>0,f(x+t)=f(x)}t0:=minT

t00,取 {tn},tiT,limntn=t0,则 f(x)=f(x+tn)=limnf(x+tn)=f(x+limntn)=f(x+t0)

t0=0,因为不是常函数, 则x0 s.t. f(x0)f(0),因为连续,和极限有保号性你可以说xO(x0,δ),f(x)f(0),但由t0=0,你可以找到t<δ,那就可以再找到N使得Nt(x0δ,x0+δ),Nt一定是周期,但f(0+Nt)f(0),矛盾.

于是得证.

Linear algebra

A fun question

In n-D space we can found at most n+1 vector v1vn+1 such that: ij,vivj<0

An example in 3-D space(the one on the book)

The construction is Obviously(CH4)

Chosen a vector, the others must be in a semisphere. We say two semisphere is seperated by plane A.

And we notice that: if two vector’s shadow on plane A construct a acute angle, their dot product must be positive, transformed it into a 2-D problems which is easy to solve.

Generalize to n-D

We choose a vector v1, and it can be written as [1,0,0](with some rotation)

so vi,i>1,viv1<0vi,1<0vi,1vj,1>0vivjvi,1vj,1<0

then transformed it into the (n-1)-D situation.

The construction can be easily give during the induction

Another Proof

given by Bing!

Lemma: Radon Partition

In n-D space, n+2 point could be divided to two convex hull with intersection.

n+2 vector must be dependent: ci s.t. i=1n+2cixi=0;ixi=0(the second condition can be satisfied by add another all-1 dimension).

so divide the vector by sign of ci we got:v=iAcixi=jBcjxj,so divide the eqution by iAci, you get one point(v) in the intersection.

We noticed 0<v2=(iAcixi)(jBcjxj)<0, contradiction!

Operator’s Left Inverse and Right Inverse

Existance

对算子T来说,左逆存在等价于右逆存在.

Proof 1

注意到左逆存在等价于T是单射,右逆存在等价于T是满射.

又因为T单射等价于T满射等价于T双射,所以左逆存在等价于右逆存在且两个逆一定相同.

Proof 2

左逆推出T分解成初等行变换矩阵,然后一个一个取逆推右逆.

Hello World

Introduction

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